Jawaban:
Penjelasan dengan langkah-langkah:
< pusat = 60°
R = 20 cm
A. Panjang busur PQ
< pusat/360 x keliling lingkaran
60/360 x 2 x 3,14 x 20
1/6 x 40 x 3,14
1/6 x 125,6
20,933 cm
20 ¹⁴/₁₅ cm
B. Luas juring OPQ
< pusat/360 x luas lingkaran
60/360 x 3,14 x 20 x 20
1/6 x 3,14 x 400
1/6 x 1.256
209,333 cm²
209 ¹/₃ cm²
Demikian
Jawab:(a.) Pjg busur ≈ 20 ¹⁴/₁₅ cm(b.) Luas juring ≈ 209 ⅓ cm²
Penjelasan:Diketahuisudut juring (θ°) = 60°jari-jari (r) = 20 cmπ ≈ 3,14
Ditanya(a.) Pjg busur= θ×2×π×r÷360≈ 60×2×3,14×20÷360≈ 6×2×3,14×20÷36≈ 12×3,14×10÷18≈ 2×3,14×10÷3≈ 3,14×20÷3≈ 62,8÷3≈ 628 ÷ 30≈ (628÷2) ÷ (30÷2)≈ 314 ÷ 15≈ (300+14)÷15≈ ³⁰⁰/₁₅ + ¹⁴/₁₅≈ 20 + ¹⁴/₁₅≈ 20 ¹⁴/₁₅ cm
(b.) Luas juring= θ×π×r²÷360≈ 60×3,14×20²÷360≈ 6×3,14×20²÷36≈ 3,14×20²÷6≈ 3,14×20×20÷6≈ 3,14×20×10÷3≈ 3,14×200÷3≈ 628÷3≈ (627+1)÷3≈ ⁶²⁷/₃ + ⅓≈ 209 + ⅓≈ 209 ⅓ cm²
(Xcvi)
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Jawaban:
[Lingkaran]
Penjelasan dengan langkah-langkah:
< pusat = 60°
R = 20 cm
A. Panjang busur PQ
< pusat/360 x keliling lingkaran
60/360 x 2 x 3,14 x 20
1/6 x 40 x 3,14
1/6 x 125,6
20,933 cm
20 ¹⁴/₁₅ cm
B. Luas juring OPQ
< pusat/360 x luas lingkaran
60/360 x 3,14 x 20 x 20
1/6 x 3,14 x 400
1/6 x 1.256
209,333 cm²
209 ¹/₃ cm²
Demikian
Jawab:
(a.) Pjg busur ≈ 20 ¹⁴/₁₅ cm
(b.) Luas juring ≈ 209 ⅓ cm²
Penjelasan:
Diketahui
sudut juring (θ°) = 60°
jari-jari (r) = 20 cm
π ≈ 3,14
Ditanya
(a.) Pjg busur
= θ×2×π×r÷360
≈ 60×2×3,14×20÷360
≈ 6×2×3,14×20÷36
≈ 12×3,14×10÷18
≈ 2×3,14×10÷3
≈ 3,14×20÷3
≈ 62,8÷3
≈ 628 ÷ 30
≈ (628÷2) ÷ (30÷2)
≈ 314 ÷ 15
≈ (300+14)÷15
≈ ³⁰⁰/₁₅ + ¹⁴/₁₅
≈ 20 + ¹⁴/₁₅
≈ 20 ¹⁴/₁₅ cm
(b.) Luas juring
= θ×π×r²÷360
≈ 60×3,14×20²÷360
≈ 6×3,14×20²÷36
≈ 3,14×20²÷6
≈ 3,14×20×20÷6
≈ 3,14×20×10÷3
≈ 3,14×200÷3
≈ 628÷3
≈ (627+1)÷3
≈ ⁶²⁷/₃ + ⅓
≈ 209 + ⅓
≈ 209 ⅓ cm²
(Xcvi)